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NCERT Exemplar · Class 12 Mathematics Probability

108 questions · 108 still being checked

EXERCISE 13.3 71–80 (part 8 of 11)

  1. Choose the correct answer from the given four options in each of the exercises from $\displaystyle 56$ to 82.

    Exercise 71

    Let A and B be two events such that P(A)=38,P(B)=58\displaystyle \mathrm{P}(\mathrm{A})=\frac{3}{8}, \mathrm{P}(\mathrm{B})=\frac{5}{8} and P(A∪B)=34\displaystyle \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}. Then P(A∣B).P(A′∣B)\displaystyle \mathrm{P}(\mathrm{A} \mid \mathrm{B}) . \mathrm{P}\left(\mathrm{A}^{\prime} \mid \mathrm{B}\right) is equal to
    (A)
    25\displaystyle \frac{2}{5}
    (B)
    38\displaystyle \frac{3}{8}
    (C)
    320\displaystyle \frac{3}{20}
    (D)
    625\displaystyle \frac{6}{25}

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    NCERT’s answer
    D
    (D) \(\displaystyle \dfrac{6}{25}\)\[P(A\cap B)=P(A)+P(B)-P(A\cup B)=\frac38+\frac58-\frac34=\frac14 \] \[P(A\mid B)=\frac{P(A\cap B)}{P(B)}=\frac{1/4}{5/8}=\frac25 \] \[P(A'\mid B)=1-P(A\mid B)=\frac35 \] \[P(A\mid B)\,P(A'\mid B)=\frac25\cdot\frac35=\frac{6}{25} \]

  2. Exercise 72

    If the events A and B are independent, then P(A∩B)\displaystyle \mathrm{P}(\mathrm{A} \cap \mathrm{B}) is equal to
    (A)
    P(A)+P(B)\displaystyle \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})
    (B)
    P(A)−P(B)\displaystyle \mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{B})
    (C)
    P(A)⋅P(B)\displaystyle \mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})
    (D)
    P(A)∣P(B)\displaystyle \mathrm{P}(\mathrm{A}) \mid \mathrm{P}(\mathrm{B})

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    NCERT’s answer
    C
    (C) \(\displaystyle P(A)\cdot P(B)\)Independence means the occurrence of one does not change the probability of the other: \[P(A\mid B)=P(A) \] \[P(A\cap B)=P(A\mid B)\,P(B)=P(A)\,P(B) \]

  3. Exercise 73

    Two events E and F are independent. If P(E)=0.3,P(E∪F)=0.5\displaystyle \mathrm{P}(\mathrm{E})=0.3, \mathrm{P}(\mathrm{E} \cup \mathrm{F})=0.5, then P(E∣F)−P(F∣E)\displaystyle \mathrm{P}(\mathrm{E} \mid \mathrm{F})-\mathrm{P}(\mathrm{F} \mid \mathrm{E}) equals
    (A)
    27\displaystyle \frac{2}{7}
    (B)
    335\displaystyle \frac{3}{35}
    (C)
    170\displaystyle \frac{1}{70}
    (D)
    17\displaystyle \frac{1}{7}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{1}{70}\). Independence gives \(\displaystyle \mathrm{P}(E\mid F)=\mathrm{P}(E)\) and \(\displaystyle \mathrm{P}(F\mid E)=\mathrm{P}(F)\).\[\mathrm{P}(E\cup F)=\mathrm{P}(E)+\mathrm{P}(F)-\mathrm{P}(E)\,\mathrm{P}(F) \]\[0.5=0.3+\mathrm{P}(F)\,(1-0.3) \]\[\mathrm{P}(F)=\frac{0.2}{0.7}=\frac{2}{7} \]\[\mathrm{P}(E\mid F)-\mathrm{P}(F\mid E)=\frac{3}{10}-\frac{2}{7}=\frac{21-20}{70}=\frac{1}{70} \]

  4. Exercise 74

    A bag contains 5\displaystyle 5 red and 3\displaystyle 3 blue balls. If 3\displaystyle 3 balls are drawn at random without replacement the probability of getting exactly one red ball is
    (A)
    45196\displaystyle \frac{45}{196}
    (B)
    135392\displaystyle \frac{135}{392}
    (C)
    1556\displaystyle \frac{15}{56}
    (D)
    1529\displaystyle \frac{15}{29}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{15}{56}\). Exactly one red means $\displaystyle 1$ red from $\displaystyle 5$ and $\displaystyle 2$ blue from 3.\[n(S)={}^{8}C_3=56 \]\[n(\text{one red})={}^{5}C_1\times{}^{3}C_2=5\times 3=15 \]\[\mathrm{P}=\frac{15}{56} \]

  5. Exercise 75

    Refer to Question 74\displaystyle 74 above. The probability that exactly two of the three balls were red, the first ball being red, is
    (A)
    13\displaystyle \frac{1}{3}
    (B)
    47\displaystyle \frac{4}{7}
    (C)
    1528\displaystyle \frac{15}{28}
    (D)
    528\displaystyle \frac{5}{28}

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    NCERT’s answer
    B
    (B) \(\displaystyle \frac{4}{7}\). Given the first ball is red, $\displaystyle 4$ red and $\displaystyle 3$ blue remain; exactly one more red must come in the next two draws.\[n(S)={}^{7}C_2=21 \]\[n(\text{one red, one blue})={}^{4}C_1\times{}^{3}C_1=12 \]\[\mathrm{P}=\frac{12}{21}=\frac{4}{7} \]

  6. Exercise 76

    Three persons, A, B and C, fire at a target in turn, starting with A. Their probability of hitting the target are 0.4,0.3\displaystyle 0.4,0.3 and 0.2\displaystyle 0.2 respectively. The probability of two hits is
    (A)
    0.024\displaystyle 024
    (B)
    0.188\displaystyle 188
    (C)
    0.336\displaystyle 336
    (D)
    0.452\displaystyle 452

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    NCERT’s answer
    B
    (B) \(\displaystyle 0.188\). Exactly two hits can happen in three ways, the shots being independent.\[\text{A, B hit; C misses: } 0.4\times0.3\times0.8=0.096 \]\[\text{A, C hit; B misses: } 0.4\times0.7\times0.2=0.056 \]\[\text{B, C hit; A misses: } 0.6\times0.3\times0.2=0.036 \]\[\mathrm{P}=0.096+0.056+0.036=0.188 \]

  7. Exercise 77

    Assume that in a family, each child is equally likely to be a boy or a girl. A family with three children is chosen at random. The probability that the eldest child is a girl given that the family has at least one girl is
    (A)
    12\displaystyle \frac{1}{2}
    (B)
    13\displaystyle \frac{1}{3}
    (C)
    23\displaystyle \frac{2}{3}
    (D)
    47\displaystyle \frac{4}{7}

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    NCERT’s answer
    D
    (D) \(\displaystyle \frac{4}{7}\). Let \(\displaystyle G\): eldest is a girl; \(\displaystyle H\): at least one girl. Eight equally likely outcomes.\[\mathrm{P}(H)=1-\mathrm{P}(\text{all boys})=1-\frac{1}{8}=\frac{7}{8} \]\[G\subset H \Rightarrow \mathrm{P}(G\cap H)=\mathrm{P}(G)=\frac{1}{2} \]\[\mathrm{P}(G\mid H)=\frac{\mathrm{P}(G\cap H)}{\mathrm{P}(H)}=\frac{1/2}{7/8}=\frac{4}{7} \]

  8. Exercise 78

    A die is thrown and a card is selected at random from a deck of 52\displaystyle 52 playing cards. The probability of getting an even number on the die and a spade card is
    (A)
    12\displaystyle \frac{1}{2}
    (B)
    14\displaystyle \frac{1}{4}
    (C)
    18\displaystyle \frac{1}{8}
    (D)
    34\displaystyle \frac{3}{4}

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    NCERT’s answer
    C
    (C) \(\displaystyle \frac{1}{8}\). The die and the card are independent.\[\mathrm{P}(\text{even})=\frac{3}{6}=\frac{1}{2},\qquad \mathrm{P}(\text{spade})=\frac{13}{52}=\frac{1}{4} \]\[\mathrm{P}(\text{both})=\frac{1}{2}\times\frac{1}{4}=\frac{1}{8} \]

  9. Exercise 79

    A box contains 3\displaystyle 3 orange balls, 3\displaystyle 3 green balls and 2\displaystyle 2 blue balls. Three balls are drawn at random from the box without replacement. The probability of drawing 2\displaystyle 2 green balls and one blue ball is
    (A)
    328\displaystyle \frac{3}{28}
    (B)
    221\displaystyle \frac{2}{21}
    (C)
    128\displaystyle \frac{1}{28}
    (D)
    167168\displaystyle \frac{167}{168}

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    NCERT’s answer
    A
    (A) \(\displaystyle \frac{3}{28}\). Choose $\displaystyle 2$ of the $\displaystyle 3$ green and $\displaystyle 1$ of the $\displaystyle 2$ blue.\[n(S)={}^{8}C_3=56 \]\[n(E)={}^{3}C_2\times{}^{2}C_1=3\times 2=6 \]\[\mathrm{P}=\frac{6}{56}=\frac{3}{28} \]

  10. Exercise 80

    A flashlight has 8\displaystyle 8 batteries out of which 3\displaystyle 3 are dead. If two batteries are selected without replacement and tested, the probability that both are dead is
    (A)
    3356\displaystyle \frac{33}{56}
    (B)
    964\displaystyle \frac{9}{64}
    (C)
    114\displaystyle \frac{1}{14}
    (D)
    328\displaystyle \frac{3}{28}

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    NCERT’s answer
    D
    (D) \(\displaystyle \frac{3}{28}\). Both batteries chosen must come from the $\displaystyle 3$ dead ones.\[n(S)={}^{8}C_2=28,\qquad n(E)={}^{3}C_2=3 \]\[\mathrm{P}=\frac{3}{28} \]Check by multiplication:\[\frac{3}{8}\times\frac{2}{7}=\frac{6}{56}=\frac{3}{28} \]